A262704 - OEIS

%I #44 Sep 08 2022 08:46:14

%S 1,0,1,0,6,1,0,6,24,1,0,0,114,60,1,0,0,180,690,120,1,0,0,90,2940,2640,

%T 210,1,0,0,0,5670,21840,7770,336,1,0,0,0,5040,87570,107520,19236,504,

%U 1,0,0,0,1680,189000,735210,407400,42084,720,1,0,0,0,0,224700,2835756,4280850,1284360,83880,990,1

%N Triangle: Newton expansion of C(n,m)^3, read by rows.

%C Triangle here T_3(n,m) is such that C(n,m)^3 = Sum_{j=0..n} C(n,j)*T_3(j,m).

%C Equivalently, lower triangular matrix T_3 such that

%C || C(n,m)^3 || = A181583 =  P * T_3 = A007318 * T_3.

%C T_3(n,m) = 0 for n < m and for 3*m < n. In fact:

%C C(x,m)^q and C(x,m), with m nonnegative and q positive integer, are polynomial in x of degree m*q and m respectively, and C(x,m) is a divisor of C(x,m)^q.

%C Therefore the Newton series will give C(x,m)^q = T_q(m,m)*C(x,m) + T_q(m+1,m)*C(x,m+1) + ... + T_q(q*m,m)*C(x,q*m), where T_q(n,m) is the n-th forward finite difference of C(x,m)^q at x = 0.

%C Example:

%C C(x,2)^3 = x^3*(x-1)^3 / 8 = 1*C(x,2) + 24*C(x,3) + 114*C(x,4) + 180*C(x,5) + 90*C(x,6);

%C C(5,2)^3 = C(5,3)^3 = 1000 = 1*C(5,2) + 24*C(5,3) + 114*C(5,4) + 180*C(5,5) = 1*C(5,3) + 60*C(5,4) + 690*C(5,5).

%C So we get the expansion of the 3rd power of the binomial coefficient in terms of the binomial coefficients on the same row.

%C T_1 is the unitary matrix,

%C T_2 is the transpose of A109983,

%C T_3 is this sequence,

%C T_4, T_5 are A262705, A262706.

%H Gheorghe Coserea, <a href="/A262704/b262704.txt">Rows n = 0..200, flattened</a>

%H P. Blasiak, K. A. Penson and A. I. Solomon, <a href="http://www.arxiv.org/abs/quant-ph/0402027">The general boson normal ordering problem</a>, arXiv:quant-ph/0402027, 2004.

%F T_3(n,m) = Sum_{j=0..n} (-1)^(n-j)*C(n,j)*C(j,m)^3.

%F Also, let S(r,s)(n,m) denote the Generalized Stirling2 numbers as defined in the link above,then T_3(n,m) = n! / (m!)^3 * S(m,m)(3,n).

%e Triangle starts:

%e n\m  [0]     [1]     [2]     [3]     [4]     [5]     [6]     [7]     [8]

%e [0]  1;

%e [1]  0,      1;

%e [2]  0,      6,      1;

%e [3]  0,      6,      24,     1;

%e [4]  0,      0,      114,    60,     1;

%e [5]  0,      0,      180,    690,    120,    1;

%e [6]  0,      0,      90,     2940,   2640,   210,    1;

%e [7]  0,      0,      0,      5670,   21840,  7770,   336,    1;

%e [8]  0,      0,      0,      5040,   87570,  107520, 19236,  504,    1;

%e [9]  ...

%t T3[n_, m_] := Sum[(-1)^(n - j) * Binomial[n, j] * Binomial[j, m]^3, {j, 0, n}]; Table[T3[n, m], {n, 0, 10}, {m, 0, n}] // Flatten (* _Jean-François Alcover_, Oct 01 2015 *)

%o (MuPAD)

%o // as a function

%o T_3:=(n,m)->_plus((-1)^(n-j)*binomial(n,j)*binomial(j,m)^3 $ j=0..n):

%o // as a matrix h x h

%o _P:=h->matrix([[binomial(n,m) $m=0..h]$n=0..h]):

%o _P_3:=h->matrix([[binomial(n,m)^3 $m=0..h]$n=0..h]):

%o _T_3:=h->_P(h)^-1*_P_3(h):

%o (PARI) T_3(nmax) = {for(n=0, nmax, for(m=0, n, print1(sum(j=0, n, (-1)^(n-j)*binomial(n,j)*binomial(j,m)^3), ", ")); print())} \\ _Colin Barker_, Oct 01 2015

%o (Magma) [&+[(-1)^(n-j)*Binomial(n,j)*Binomial(j,m)^3: j in [0..n]]: m in [0..n], n in [0..10]]; // _Bruno Berselli_, Oct 01 2015

%o (PARI) t3(n,m) = sum(j=0, n,  (-1)^((n-j)%2)* binomial(n,j)*binomial(j,m)^3);

%o concat(vector(11, n, vector(n, k, t3(n-1,k-1)))) \\ _Gheorghe Coserea_, Jul 14 2016

%Y Row sums are A172634, the inverse binomial transform of the Franel numbers (A000172).

%Y Column sums are the A126086, per the comment given thereto by _Brendan McKay_.

%Y Second diagonal (T_3(n+1,n)) is A007531 (n+2).

%Y Column T_3(n,2) is A122193(3,n).

%Y Cf. A109983 (transpose of), A262705, A262706.

%Y Cf. A078739, A078741, A274786.

%K nonn,tabl,easy

%O 0,5

%A _Giuliano Cabrele_, Sep 27 2015